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Chapter 1: Problem 11

Use Mathematical Induction to prove that if the set \(S\) has \(n\) elements, then \(\mathcal{P}(S)\) has \(2^{n}\) elements.

Short Answer

Expert verified
Using mathematical induction, it is proven that if a set S has n elements then its power set \( \mathcal{P}(S) \) has \( 2^{n} \) elements.

Step by step solution

01

Base Case

First, checking the base case when \(n = 0\). The set \(S\) has no elements, i.e., \(S = \emptyset\). The power set \(\mathcal{P}(S)\) of \(\emptyset\) includes only the empty set and has \(1 = 2^{0}\) elements. So the theorem holds true for \(n = 0\).
02

Induction Step - Assume True for n

Next, assuming the theorem is true for \(n = k\). That is, a set \(S\) with \(k\) elements has a power set with \(2^{k}\) elements.
03

Induction Step - Prove True for n+1

Now, proving the theorem for \(n = k + 1\). If a set \(T\) has \(k + 1\) elements, one of which is \(x\), then \(T = S \cup \{x\}\), where \(S\) is a subset with \(k\) elements. The power set \(\mathcal{P}(T)\) consists of \(\mathcal{P}(S)\) and sets from \(\mathcal{P}(S)\) with \(x\). Both are distinct and have \(2^{k}\) elements each. Consequently, \(\mathcal{P}(T)\) has \(2*2^{k} = 2^{k+1}\) elements. This completes the induction.

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